A certain species of bacteria grows at a constant rate, its numbers in

Question

A certain species of bacteria grows at a constant rate, its numbers increasing by an unknown multiplier every hour. At the end of one hour (time  t  = 1), the bacteria colony’s population size is  p ; at the end of five hours ( t  = 5), the colony’s population has grown to  p^2 .

In the first column, indicate the factor by which the number of bacteria grows each hour. In the second column, indicate the population at the start of the time period described (at time t = 0). Make only two selections, one in each column.

mikemcgarry · Accepted Answer

Dear AccipiterQ , I'm happy to help. I believe the OA given is correct, and does correctly account for exponential growth. First, here's a blog on exponents: https://magoosh.com/gmat/2012/exponent-p ... -the-gmat/ Part of what's tricky is that the start time, with population p, is at t = 1, and the the later time, at t = 5, is FOUR hours later. That's the first tricky thing. Also, keep in mind, for exponential growth, of course the size of the population doesn't increase linearly, but the exponent on the function does increase linearly . Thus, at time #1, the population is a P^A, and at time #2, it's at P^B (where B>A), then at a time exactly between those two times, the population will have an exponent exactly between those two values (a.k.a. the average), P^ Part one: Let r be the unknown hourly growth ratio: t = 1 --> p t = 2 --> p*r t = 3 --> p*r*r = p*(r^2) t = 4 --> p*r*r*r = p*(r^3) t = 5 --> p*r*r*r*r = p*(r^4) Set that last value equal to p^2 p*(r^4) = p^2 r^4 = p r = p^(1/4) Column #1, choice = (A) Given that t = 1 --> p t = 0 --> p/r = p/ = p^(3/4) Column #2, choice = (D) Does all this make sense? Mike :-)

panthersarkar · Answer

Let us consider r the factor.

At end of 1 hr......p
At end of 2 hr......pr
.
.
At end of 5 hr....pr^4

pr^4=p^2

r=p^1/4

At t=0 , P^(1-1/4)=p^3/4 ...Ans

rbatista19 · Answer

Another way to see this: p^2 = p^(5t+t0)
By experimenting several values we get to p(t0)=p^3/4 & GF = p^1/4

Aashimabhatia · Answer

Let n be the growth factor
1st Year- P
2nd Year- Pn
3rd Year- Pn^2
4th Year- Pn^3
5th Year- Pn^4
This means P^2=Pn^4
P=n^4
n= P^1/4
When t=0
Population = P/n = P/P^1/4 = P^3/4

ku2002 · Answer

If you observe, it is in GP form.

We can just get the answer by eliminating option.

let k be the unknown constant rate, a0 be the population in the initial state at t=0 and a1 be the population at t=1 which is a1=p.

we know as per GP, a1/a0 = k => a1=a0*k p=a0*k, p=p^3/4 * p^1/4

Growth Factor	Population at t = 0
		\(p^\frac{1}{4}\)
		\(p^\frac{1}{3}\)
		\(p^\frac{1}{2}\)
		\(p^\frac{3}{4}\)
		\(p\)
		\(p^\frac{3}{2}\)

A certain species of bacteria grows at a constant rate, its numbers in

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